Given a right angle and the equations of two sides, can we find the third?

Given a right angle and the equations of two sides, can we find the third?

Question

In the triangle $ABC$ the equations of two of the sides are $2y=x+2$ and $y=2x-2$. Given that $A$ is the point $(4,3)$ and that $A\hat{B}C$ is $90^\circ$, find the equation of the third side.

Suggestion

How could we find out which of those sides goes through $A$?

Solution

Our best starting plan is to draw a diagram.

The point $A$ must be on either $y = 2x-2$, or $2y = x + 2$, and we can see by substituting in that $A$ is on the second of these but not the first.

So the first line ($y=2x-2$) must give the side $BC$, so $B$ lies on this line. We are also told that the angle at $B$ is a right angle.

We can now sketch our diagram. (At this stage, we have not worked out whether $2y=x+2$ passes through $B$ or not.)

Now the gradient of $BC$ is $2$, so the gradient of $AB$ is $-\frac12$, since $AB$ is perpendicular to $BC$. Since the line $AB$ passes through $A(4,3)$, it follows that it has equation $y-3=-\frac12(x-4)$ (using the general formula $y-y_1=m(x-x_1)$), which rearranges to $y=-\frac12x+5$ or $x+2y=10$. This is the equation of the third side.

It also follows that $2y=x+2$ is the line $AC$ (since it has gradient $+\frac12$, so is not perpendicular to $BC$). This would allow us to work out the coordinates of $C$, though we are not asked to find these.